If all roots of the equation x^3 - 2ax^2 + 3b | vedclass

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Question

(A) Let the roots of the equation be $x_1, x_2, x_3 > 0$.From Vieta's formulas:$x_1 + x_2 + x_3 = 2a$$x_1x_2 + x_2x_3 + x_3x_1 = 3b$$x_1x_2x_3 = 8$Usi

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let the roots of the equation be $x_1, x_2, x_3 > 0$.From Vieta's formulas:$x_1 + x_2 + x_3 = 2a$$x_1x_2 + x_2x_3 + x_3x_1 = 3b$$x_1x_2x_3 = 8$Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality for the terms $x_1x_2, x_2x_3, x_3x_1$:$\frac{x_1x_2 + x_2x_3 + x_3x_1}{3} \geq \sqrt[3]{(x_1x_2)(x_2x_3)(x_3x_1)}$$\frac{3b}{3} \geq \sqrt[3]{(x_1x_2x_3)^2}$$b \geq \sqrt[3]{8^2}$$b \geq \sqrt[3]{64}$$b \geq 4$Therefore,the minimum value of $b$ is $4$.

If all roots of the equation $x^3 - 2ax^2 + 3bx - 8 = 0$ are positive,where $a, b \in R$,then the minimum value of $b$ is:

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